Rename this forum Voodoo Betting Strategies?
- Thread starter zengrifter
- Start date
Voodoo or logic
Because a few well known betting strategies have been thoroughly analyzed and proven not to work someone jumps to the conclusion that all unknown ones don't work either. That kind of logic is a better example of voodooism than any betting system could possibly be. The ones that work arn't likely to appear on any open forum for the casinos to read about.
Understanding why a particular betting system doesn't work is more important to the player than calling it voodoo.
Because a few well known betting strategies have been thoroughly analyzed and proven not to work someone jumps to the conclusion that all unknown ones don't work either. That kind of logic is a better example of voodooism than any betting system could possibly be. The ones that work arn't likely to appear on any open forum for the casinos to read about.
Understanding why a particular betting system doesn't work is more important to the player than calling it voodoo.
Everytime you win a hand you have overcome the house edge.
Proof is a very strong work. They have shown that counting works long term.
Everytime a counter increases his bet when the count is high, and loses, is proof that nothing has been proved. Everytime a non-counter wins a hand he has overcome the house edge in that instance. Bj is a very close game and it would be difficult to predict with certainty if the next hand will win or lose.
Counters will use s.d. and variance to explain away the holes in their proof but frown if non-counters use the same arguments. Counting doesn't overcome the house edge, it just takes advantage of a temporary house weakness. In some ways non-counters have an advantage over counters, at the beginning of a shoe, because the counters have to wait to get the info they need to decide what to bet.
Their mistake is in making too many assumptions about progressions. Another is in trusting high counts that may be drastically affected by the cards behind the cut card.
Proof is a very strong work. They have shown that counting works long term.
Everytime a counter increases his bet when the count is high, and loses, is proof that nothing has been proved. Everytime a non-counter wins a hand he has overcome the house edge in that instance. Bj is a very close game and it would be difficult to predict with certainty if the next hand will win or lose.
Counters will use s.d. and variance to explain away the holes in their proof but frown if non-counters use the same arguments. Counting doesn't overcome the house edge, it just takes advantage of a temporary house weakness. In some ways non-counters have an advantage over counters, at the beginning of a shoe, because the counters have to wait to get the info they need to decide what to bet.
Their mistake is in making too many assumptions about progressions. Another is in trusting high counts that may be drastically affected by the cards behind the cut card.
Yes, it is a very strong work.
http://www.bjmath.com/bjmath/progress/unfair.htm (Archive copy)
In "The Casino Gambler's Guide," Allan Wilson provided a mathematical proof of the fallacy that a progression can overcome a negative expectation in a game with even payoffs. This article expands on Wilson's Proof and provides the proof that progression systems cannot overcome a negative expectation even if the game provides uneven payoffs.
Let bk = the size of the kth bet.
Mk = the size of the payoff on the kth bet.
pk = the probability that the series terminates with a win on the kth bet, having been preceeded by k-1 losses in a row.
n-1 = the greatest number of losses in a row that a player can handle, given the size of the player's bankroll. In other words, the nth bet must be won, otherwise the player's entire bankroll will be lost.
Let's now define Bn = bn * Mn
The expected value for any series is:
Eseries = p1B1 + p2(B2-b1) + p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2- . . . -b2-b1) + (1-p1-p2- . . . -pn) * (-bn-bn-1-bn-2- . . . -b2-b1)
If we let
Eseries = A + B where,
A = p1B1 + p2(B2-b1) + p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2- . . . -b2-b1)
and
B = (1-p1-p2- . . . -pn) * (-bn-bn-1-bn-2- . . . -b2-b1)
then it is easier to see that "A" represents the probability that the series will end with a win multiplied by the bet size at the nth term in the series and "B" is the probability that the series ends in a loss multiplied by the net loss.
Now let's rearrange the terms in "A."
A = p1B1 + p2B2 - p2b1 + p3B3 - p3b2 - p3b1 + . . . + pnBn - pnbn-1 - pnbn-2 - . . . - pnb2 - pnb1
A = p1B1 + p2B2 + . . . + pnBn + b1(- p2 - p3 - . . . - pn) + b2(- p3 - . . . - pn) + bn-2(- pn-1 - pn) + bn-1(- pn)
And for "B" we get
B = -bn(1 - p1 - p2 - . . . - pn) - bn-1(1 - p1 - p2 - . . . - pn) - . . . - b2(1 - p1 - p2 - . . . - pn) - b1(1 - p1 - p2 - . . . - pn)
Now if we combine A and B again, we get,
Eseries = A + B
Eseries = p1B1 + p2B2 + . . . + pnBn - b1(1 - p1) - b2(1 - p1 - p2) - . . . - bn-1(1 - p1 - p2 - . . . - pn-1) - bn(1 - p1 - p2 - . . . - pn)
Eseries = p1B1 + p2B2 + . . . + pnBn + p1b1 + (p2 + p1)b2 + (p3 + p2 + p1)b3 + . . . + (pn + pn-1 + . . . + p2 + p1)bn - (b1 + b2 + . . . + bn)
Wilson points out that to get rid of the subscripts, all we have to do is realize that pk = (1-p)k-1p, where p is the probability of a win on an individual play and 1-p is the probability of a loss. If we think about it, it makes sense that the probability of a series terminating in a win at the kth level is the product of the probability of k-1 losses in a row multiplied by the probability of win on the kth trial.
So how do we use this information? Well, let's try substituting this expression for each pk and see what we get.
Eseries = (1-p)1-1pB1 + (1-p)2-1pB2 + . . . + (1-p)n-1pBn + (1-p)1-1pb1 + ((1-p)2-1p + (1-p)1-1p)b2 + ((1-p)3-1p + (1-p)2-1p + (1-p)1-1p)b3 + . . . + ((1-p)n-1p + (1-p)n-1-1p + . . . + (1-p)2-1p + (1-p)1-1p)bn - (b1 + b2 + . . . + bn)
Simplifying, we get
Eseries = (p(1-p)0B1 + (1-p)1pB2 + . . . + (1-p)n-1pBn + p(1-p)0b1 + ((1-p)1p + (1-p)0p)b2 + ((1-p)2p + (1-p)1p + (1-p)0p)b3 + . . . + ((1-p)n-1p + (1-p)n-2p + . . . + (1-p)1p + (1-p)0p)bn - (b1 + b2 + . . . + bn)
If we factor p out of the first parts of the equation and look closely, we can see that the kth term T can be written as:
T = p[(1-p)k-1]Bk + p[(1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
or rephrased for Bk = bk * Mk we get
T = p[(1-p)k-1Mk + (1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
If we substitute
C = (1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0
and if we multiply C by (1-p) and call this D
D = (1-p)C = (1-p)k + (1-p)k-1 + . . . + (1-p)3 + (1-p)2 + (1-p)1
Now, if we subtract C from D, we get
D - C = (1-p)C - C = (1-p)k - (1-p)0
[(1-p) - 1]C = (1-p)k - (1-p)0
C = [(1-p)k - 1]/[(1-p) - 1] or
C = [(1-p)k - 1]/-p]
Now if we substitute C back into T, we get
T = p[(1-p)k-1Mk + [(1-p)k - 1]/-p]bk
T = p(1-p)k-1Mk + 1 - (1-p)k]bk
T = p(1-p)k-1Mk + 1 - (1-p)(1-p)k-1]bk
T = [[pMk - (1-p)](1-p)k-1 + 1]bk This now allows us to write the equations in terms of summations. We therefore get
Eseries = sum {[[pMk - (1-p)](1-p)k-1]bk} + sum {bk} - sum {bk}, for k = 1 to n
The last two terms cancel, so we are left with:
Eseries = sum {[pMk - (1-p)](1-p)k-1]bk}, for k = 1 to n
Eseries = sum {[(1+Mk)p - 1](1-p)k-1]]bk}, for k = 1 to n
If we now look closely at this equation, we can make several observations. First, the sign of Eseries depends solely on the resulting sign of [(1+Mk)p - 1]. To make things a little easier to follow, let's say we're dealing with a game that has even payoffs. This means that Mk = 1 and therefore
Eseries = sum {[2p - 1](1-p)k-1]]bk}, for k = 1 to n
Eseries = [2p - 1]sum {(1-p)k-1]]bk}, for k = 1 to n
Now it is a little easier to see what is going on. For example, if we are in an unfair game, then p < 0.5 and we can easily see that 2p-1 will be a negative value. For example, if our chance of winning is only 49%, then p = 0.49 and 2p-1 = -0.02. In an even game, p = 0.5 and we see that 2*0.5-1 = 0. In this case, the equation is telling us that in an even game the expected value is zero just as we would expect it should. If we are playing a game with an advantage, then p > 0.5 and 2p-1 will be positive.
The general formula for uneven payoffs work just as well, but is more complicated to understand. Suffice to say that if [(1+Mk)p - 1] is negative, then regardless of the progression, the game will eventually result in a loss for the player.
Hopefully, this post will provide definitive proof of the fallacy of trying to overcome a negative expectation by using any type of progression whether it be the martingale or some other modern progression.
http://www.bjmath.com/bjmath/progress/unfair.htm (Archive copy)
In "The Casino Gambler's Guide," Allan Wilson provided a mathematical proof of the fallacy that a progression can overcome a negative expectation in a game with even payoffs. This article expands on Wilson's Proof and provides the proof that progression systems cannot overcome a negative expectation even if the game provides uneven payoffs.
Let bk = the size of the kth bet.
Mk = the size of the payoff on the kth bet.
pk = the probability that the series terminates with a win on the kth bet, having been preceeded by k-1 losses in a row.
n-1 = the greatest number of losses in a row that a player can handle, given the size of the player's bankroll. In other words, the nth bet must be won, otherwise the player's entire bankroll will be lost.
Let's now define Bn = bn * Mn
The expected value for any series is:
Eseries = p1B1 + p2(B2-b1) + p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2- . . . -b2-b1) + (1-p1-p2- . . . -pn) * (-bn-bn-1-bn-2- . . . -b2-b1)
If we let
Eseries = A + B where,
A = p1B1 + p2(B2-b1) + p3(B3-b2-b1) + . . . + pn(Bn-bn-1-bn-2- . . . -b2-b1)
and
B = (1-p1-p2- . . . -pn) * (-bn-bn-1-bn-2- . . . -b2-b1)
then it is easier to see that "A" represents the probability that the series will end with a win multiplied by the bet size at the nth term in the series and "B" is the probability that the series ends in a loss multiplied by the net loss.
Now let's rearrange the terms in "A."
A = p1B1 + p2B2 - p2b1 + p3B3 - p3b2 - p3b1 + . . . + pnBn - pnbn-1 - pnbn-2 - . . . - pnb2 - pnb1
A = p1B1 + p2B2 + . . . + pnBn + b1(- p2 - p3 - . . . - pn) + b2(- p3 - . . . - pn) + bn-2(- pn-1 - pn) + bn-1(- pn)
And for "B" we get
B = -bn(1 - p1 - p2 - . . . - pn) - bn-1(1 - p1 - p2 - . . . - pn) - . . . - b2(1 - p1 - p2 - . . . - pn) - b1(1 - p1 - p2 - . . . - pn)
Now if we combine A and B again, we get,
Eseries = A + B
Eseries = p1B1 + p2B2 + . . . + pnBn - b1(1 - p1) - b2(1 - p1 - p2) - . . . - bn-1(1 - p1 - p2 - . . . - pn-1) - bn(1 - p1 - p2 - . . . - pn)
Eseries = p1B1 + p2B2 + . . . + pnBn + p1b1 + (p2 + p1)b2 + (p3 + p2 + p1)b3 + . . . + (pn + pn-1 + . . . + p2 + p1)bn - (b1 + b2 + . . . + bn)
Wilson points out that to get rid of the subscripts, all we have to do is realize that pk = (1-p)k-1p, where p is the probability of a win on an individual play and 1-p is the probability of a loss. If we think about it, it makes sense that the probability of a series terminating in a win at the kth level is the product of the probability of k-1 losses in a row multiplied by the probability of win on the kth trial.
So how do we use this information? Well, let's try substituting this expression for each pk and see what we get.
Eseries = (1-p)1-1pB1 + (1-p)2-1pB2 + . . . + (1-p)n-1pBn + (1-p)1-1pb1 + ((1-p)2-1p + (1-p)1-1p)b2 + ((1-p)3-1p + (1-p)2-1p + (1-p)1-1p)b3 + . . . + ((1-p)n-1p + (1-p)n-1-1p + . . . + (1-p)2-1p + (1-p)1-1p)bn - (b1 + b2 + . . . + bn)
Simplifying, we get
Eseries = (p(1-p)0B1 + (1-p)1pB2 + . . . + (1-p)n-1pBn + p(1-p)0b1 + ((1-p)1p + (1-p)0p)b2 + ((1-p)2p + (1-p)1p + (1-p)0p)b3 + . . . + ((1-p)n-1p + (1-p)n-2p + . . . + (1-p)1p + (1-p)0p)bn - (b1 + b2 + . . . + bn)
If we factor p out of the first parts of the equation and look closely, we can see that the kth term T can be written as:
T = p[(1-p)k-1]Bk + p[(1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
or rephrased for Bk = bk * Mk we get
T = p[(1-p)k-1Mk + (1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0]bk
If we substitute
C = (1-p)k-1 + (1-p)k-2 + . . . + (1-p)2 + (1-p)1 + (1-p)0
and if we multiply C by (1-p) and call this D
D = (1-p)C = (1-p)k + (1-p)k-1 + . . . + (1-p)3 + (1-p)2 + (1-p)1
Now, if we subtract C from D, we get
D - C = (1-p)C - C = (1-p)k - (1-p)0
[(1-p) - 1]C = (1-p)k - (1-p)0
C = [(1-p)k - 1]/[(1-p) - 1] or
C = [(1-p)k - 1]/-p]
Now if we substitute C back into T, we get
T = p[(1-p)k-1Mk + [(1-p)k - 1]/-p]bk
T = p(1-p)k-1Mk + 1 - (1-p)k]bk
T = p(1-p)k-1Mk + 1 - (1-p)(1-p)k-1]bk
T = [[pMk - (1-p)](1-p)k-1 + 1]bk This now allows us to write the equations in terms of summations. We therefore get
Eseries = sum {[[pMk - (1-p)](1-p)k-1]bk} + sum {bk} - sum {bk}, for k = 1 to n
The last two terms cancel, so we are left with:
Eseries = sum {[pMk - (1-p)](1-p)k-1]bk}, for k = 1 to n
Eseries = sum {[(1+Mk)p - 1](1-p)k-1]]bk}, for k = 1 to n
If we now look closely at this equation, we can make several observations. First, the sign of Eseries depends solely on the resulting sign of [(1+Mk)p - 1]. To make things a little easier to follow, let's say we're dealing with a game that has even payoffs. This means that Mk = 1 and therefore
Eseries = sum {[2p - 1](1-p)k-1]]bk}, for k = 1 to n
Eseries = [2p - 1]sum {(1-p)k-1]]bk}, for k = 1 to n
Now it is a little easier to see what is going on. For example, if we are in an unfair game, then p < 0.5 and we can easily see that 2p-1 will be a negative value. For example, if our chance of winning is only 49%, then p = 0.49 and 2p-1 = -0.02. In an even game, p = 0.5 and we see that 2*0.5-1 = 0. In this case, the equation is telling us that in an even game the expected value is zero just as we would expect it should. If we are playing a game with an advantage, then p > 0.5 and 2p-1 will be positive.
The general formula for uneven payoffs work just as well, but is more complicated to understand. Suffice to say that if [(1+Mk)p - 1] is negative, then regardless of the progression, the game will eventually result in a loss for the player.
Hopefully, this post will provide definitive proof of the fallacy of trying to overcome a negative expectation by using any type of progression whether it be the martingale or some other modern progression.
The error is always at the beginning.
Where does he make allowances for winnings?
Anyone who loses umpteen hands in a row is going to lose nomatter who you are. That's just plain logic.
A progression player is going to lose some sessions, so is everyone else.
Maybe the progression is going to use Kelly betting to conserve his bankroll.
He assumes the person is going to go all the way. Even a counter is going to lose his bankroll if he goes all the way.
Winnings do affect how a person continues.
Where does he make allowances for winnings?
Anyone who loses umpteen hands in a row is going to lose nomatter who you are. That's just plain logic.
A progression player is going to lose some sessions, so is everyone else.
Maybe the progression is going to use Kelly betting to conserve his bankroll.
He assumes the person is going to go all the way. Even a counter is going to lose his bankroll if he goes all the way.
Winnings do affect how a person continues.
we agree it has been shown that counting works long term. but lets add to that. lets also agree that a card counter could lose his entire bankroll. can we also agree that a card counter is gambling with an advantage of circa 1% ?
right this is short term anything can happen in the short term. momentary events are relatively meaningless with respect to the long term statistical advantage that a card counter has.
would it be fair to say that if a non-counter wins a hand that such a win is a short term event and just as statisticaly meaningless as when a card counter loses a hand on a big bet with a high count? thereby nothing is proved once again. so really the house edge is not overcome from the long term perspective when a non-counter wins a hand it is just a relatively meaningless short term fluctuation.
not really jomo. counters use s.d. and variance as tools of the trade inorder to hone their edge and protect themselves against the gamble that they realize they are taking. it would be nothing wrong for non-counters to apply the same tools to their methods.
actually counting if done correctly can afford the counter a larger advantage over the house than the house realizes over the counter in the long term.
and yes counting does take advantage of temporary house weakness's.
but you are ignoring the disadvantages the non-counters also face at the begining of the shoe. the non-counters are are just as clueless as the counters in the begining of the shoe.
a card counter worth his salt shall have tailored his overall tactics to account for the effect of advantageous cards trapped behind the cut card.
a player who simply throws money in various ways at a game that has a long term advantage against him is the one making to many assumptions.
the fundamental differance between the way a card counter bets and the way a non-cardcounting progressionist bets is that the card counter tailors his bets according to intelligence (with regard to relative advantage or disadvantage) that he is able to glean from knowing the count where as the non-cardcounting progressionist bets according to a system of increasing and decreasing his bets but with out regard to if he has or doesn't have an advantage. both types of players are gambling. one faces the peril with a disadvantage the other with an advantage. one uses intelligence to make betting decisions. the other does not. it is entirely possible that the counter could lose his entire bankroll and one day quit in defeat while a non-counter may increase his bankroll and play on and on winning more and more. thats whats possible. here is what is likely, the card counter shall in the long run see his bankroll increase where as the non-card counter shall in the long run see his bankroll disapear.
best regards,
mr fr0g
right this is short term anything can happen in the short term. momentary events are relatively meaningless with respect to the long term statistical advantage that a card counter has.
would it be fair to say that if a non-counter wins a hand that such a win is a short term event and just as statisticaly meaningless as when a card counter loses a hand on a big bet with a high count? thereby nothing is proved once again. so really the house edge is not overcome from the long term perspective when a non-counter wins a hand it is just a relatively meaningless short term fluctuation.
not really jomo. counters use s.d. and variance as tools of the trade inorder to hone their edge and protect themselves against the gamble that they realize they are taking. it would be nothing wrong for non-counters to apply the same tools to their methods.
actually counting if done correctly can afford the counter a larger advantage over the house than the house realizes over the counter in the long term.
and yes counting does take advantage of temporary house weakness's.
but you are ignoring the disadvantages the non-counters also face at the begining of the shoe. the non-counters are are just as clueless as the counters in the begining of the shoe.
a card counter worth his salt shall have tailored his overall tactics to account for the effect of advantageous cards trapped behind the cut card.
a player who simply throws money in various ways at a game that has a long term advantage against him is the one making to many assumptions.
the fundamental differance between the way a card counter bets and the way a non-cardcounting progressionist bets is that the card counter tailors his bets according to intelligence (with regard to relative advantage or disadvantage) that he is able to glean from knowing the count where as the non-cardcounting progressionist bets according to a system of increasing and decreasing his bets but with out regard to if he has or doesn't have an advantage. both types of players are gambling. one faces the peril with a disadvantage the other with an advantage. one uses intelligence to make betting decisions. the other does not. it is entirely possible that the counter could lose his entire bankroll and one day quit in defeat while a non-counter may increase his bankroll and play on and on winning more and more. thats whats possible. here is what is likely, the card counter shall in the long run see his bankroll increase where as the non-card counter shall in the long run see his bankroll disapear.
best regards,
mr fr0g
These posts echo the prevailing view.
It's to my advantage to have that view prevail. Counters talking about counting can speak with good authority. When they speak outside their realm and comment on progressions they are on shaky ground. I understand counting and why it works and it helps me to better understand progressions, which is my main focus. In many ways counters are partly responsible for my progression and in this case the student has passed the teachers.
It's to my advantage to have that view prevail. Counters talking about counting can speak with good authority. When they speak outside their realm and comment on progressions they are on shaky ground. I understand counting and why it works and it helps me to better understand progressions, which is my main focus. In many ways counters are partly responsible for my progression and in this case the student has passed the teachers.
If you are looking for a new betting strategy:
This has worked well for me producing 3 straight session wins each $100 plus: Play with $500 session bankroll starting with $5 bets. Continue this betting cycle throughout, do not start over on the new shoe. Here it is: First bet $5, then up ONE unit on a LOSS, down TWO units on a WIN. You can average more losses than wins as you normally do in BJ and still produce a session win with this betting cycle. The most I have been up to is $50 bets, but with the down two units on a win you will eventually return to minimum bets. Do not go beyond $50 maximum bet on a $500 session bankroll starting at $5 minimum bets. Give it a try if you are losing on your current betting strategy, and hopefully you will have the same results..
This has worked well for me producing 3 straight session wins each $100 plus: Play with $500 session bankroll starting with $5 bets. Continue this betting cycle throughout, do not start over on the new shoe. Here it is: First bet $5, then up ONE unit on a LOSS, down TWO units on a WIN. You can average more losses than wins as you normally do in BJ and still produce a session win with this betting cycle. The most I have been up to is $50 bets, but with the down two units on a win you will eventually return to minimum bets. Do not go beyond $50 maximum bet on a $500 session bankroll starting at $5 minimum bets. Give it a try if you are losing on your current betting strategy, and hopefully you will have the same results..